Learn work definition & formula in physics, practice positive and negative work equation. Practice for calculating work done formula with example, dot product of two vectors and angle between vectors.

- Work is defined as the amount of energy transferred to a body by application of force.
- Work is done on an object only when there is force applied on it.
- If more than one force is acting on the body, then work done by different forces are to be calculated separately.
- Work done by a constant force is calculated as

\(W_F = \vec F.\vec s\)

\(= F\,s \,cos \,\theta\)

where F= Force applied

s = displacement from the point at which force is applied

\(\theta =\) Angle between \(\vec F\) and \(\vec s\)

A \(Fs\)

B \(-Fs\)

C \(\dfrac{Fs}{2}\)

D \(2\,Fs\)

- Work done by a force is the dot product of force and displacement from point of its application.

- Consider two vectors

\(\vec A = x_1 \hat i + y _1 \, \hat j +z_1 \,\hat k\)

\(\vec B = x_2 \hat i + y _2 \, \hat j +z_2 \,\hat k\)

\(\vec A . \vec B = ( x_1 \hat i + y _1 \, \hat j +z_1 \,\hat k).(x_2 \hat i + y _2 \, \hat j +z_2 \,\hat k)\)

\(= x_1 \,x_2 +y_1\,y_2 + z_1\,z_2\)

- \(\vec A . \vec B \) is a scalar quantity

Consider, a particle is displaced from its position vector

\(\vec r_1 = (x_1 \,\hat i +y_1 \,\hat j +\,z_1\, \hat k)\) to

\(\vec r_2 = (x_2 \,\hat i +y_2 \,\hat j +\,z_2\, \hat k)\)

Displacement

\(\Rightarrow \Delta \vec r = \vec r_2 -\vec r_1\)

\(\Rightarrow (x_2 - x_1 )\hat i + (y_2 - y_1 )\hat j + (z_2 - z_1 )\hat k \)

\(\Rightarrow \Delta \vec r = \Delta x\,\hat i +\Delta y\,\hat j +\Delta z\,\hat k \)

A \(10 \,J\)

B \(5 \,J\)

C \(12 \,J\)

D \(8 \,J\)

- Work done by a force may be positive, negative and also zero.

When there is no displacement from point of application.

- As work done, \(W = |\vec F| |\vec s| \,cos\,\theta\)

\(= |\vec F| \, cos\,\theta × 0\)

\(W=0\)

**Example : ** A child pushing a wall.

Force is perpendicular to displacement

As work done, \(W= |\vec F| |\vec s| \,cos\,\theta\)

\(W= |\vec F| |\vec s| \,cos\,90°\)

\(W= 0\)

**Example :** The work done by the normal force as well as the work done by the gravitational force on the car both are zero, because both forces are perpendicular to the displacement and have zero components along an axis in the direction of \(\vec s\).

Here the work done by gravitational force is zero.

When force is zero.

As \(W= |\vec F| |\vec s| \,cos\,\theta\)

\(W= 0× |\vec s| \,cos\,\theta\)

\(W= 0\)

**Example :**

- A man jumps to a height 'h'. During the flight, the work done by normal contact force is zero.
- Because when contact is lost, normal force is zero.

A \(\vec F_3\)

B \(\vec F_1\)

C \(\vec F_2\)

D \(\vec F_4\)

A \(\vec F _1, \vec F_2 \,\,{\text{Positive Work}}\) \(F_4 , \,F_5 \,\,{\text {Negative Work}}\)

B \(\vec F _1, \vec F_3 \,\,{\text{Positive Work}}\) \(\vec F_4 , \,\vec F_5 \,\,{\text {Zero Work}}\)

C \(\vec F _1, \vec F_4 \,\,{\text{Zero Work}}\) \(\vec F_2 , \,\vec F_3 \,\,{\text {Negative Work}}\)

D \(\vec F _2, \vec F_5 \,\,{\text{Positive Work}}\) \(\vec F_1, \vec F_4 , \,\vec F_3 \,\,{\text {Negative Work}}\)

- Work done by a force may be positive, negative or zero depending upon the angle between force and displacement vector.

Consider a force \(\vec F\) is applied on a body making an angle \(\theta\) with the displacement vector, as shown in figure.

\( W = \vec F_1 . \vec s_1\)

\(W= |\vec F_1||\vec s_1| \, cos\,\theta \)

- For work to be positive, \(cos\,\theta\) must be positive.

\(\therefore\) \( 0\leq \theta<90°\)

If angle between applied force and displacement vector is less then 90°, then work done by the applied force is always positive.

A \(\vec F_1\)

B \(\vec F_2\)

C \(\vec F_3\)

D \(\vec F_4\)

- As \(W=\vec F. \vec s\)

\(W= |\vec F| |\vec s| \,cos \,\theta\)

- For work to be negative \(\theta\) must be greater than 90 °

\(i.e. \, \theta > 90 ° \)

A \(\vec F_1\)

B \(\vec F_4\)

C \(\vec F_2 \,\,{\text {and }}\,\,\vec F_3\)

D \(\vec F_1 \,\,{\text {and }}\,\,\vec F_4\)